A single 50 g ice cube is dropped into a thermally insulated container holding 2

Question

A single 50 g ice cube is dropped into a thermally insulated container holding 200 g of water. The water is initially at 25oC and the ice is initially at -15 oC.

1)What is the final temperature of the system after is has come to thermal equilibrium ?

TFinal = 2.5155 oC

2)In terms of mass, how much of the ice has melted?

mmelted = 50 grams

3)Now lets drop a second 50 gram cube of ice into the system. What is the final temperature of the system after it has come to thermal equilibrium for this second time?

TFinal = 0 oC

4)What is the total mass of ice (including the first cube) melted during this entire process?

mmelted =_____________grams

Explanation / Answer

mass of ice = 50 gm = 0.005 kg

mass of water = 200 g = 0.2 kg

specific heat of water = 4186 J/kg-K

specific heat of ice = 2105 J/kg-K
latent heat of fusion = 334 kJ/kg

so considering all the ice got melted into water, final temperature be T.

heat lost by water = 0.2 * 4186 * (298 - T)

heat gained by ice = 0.005 * 2105 * 15 + 0.005 * 334000 + 0.005 * 4186 * T

these are equal, so by equallign and solving equation we get

a) T = 275.5155 K = 2.5155 degree C

b) So mass of ice melted = 50 gms (whole of it)

c) so we have 250 gms of water at 2.5155 degree Celsius and 50gm cube of ice is dropped. again same principle. since the temperature difference is nto much, so we can't expect the whole of ice to be melted.

let m of ice is melted and final temperature will be 0 degrees.

heat lost by water = 0.25 * 4186 * 2.5155

heat gained by ice = 0.005 * 2105 * 15 + m * 334000

on equalling both of them, we get m = 7.409 * 10 ^ -3 kg = 7.409 gm

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